Solve for $r$, $ \dfrac{3r - 2}{2r + 4} = \dfrac{1}{4r + 8} + \dfrac{10}{2r + 4} $
First we need to find a common denominator for all the expressions. This means finding the least common multiple of $2r + 4$ $4r + 8$ and $2r + 4$ The common denominator is $4r + 8$ To get $4r + 8$ in the denominator of the first term, multiply it by $\frac{2}{2}$ $ \dfrac{3r - 2}{2r + 4} \times \dfrac{2}{2} = \dfrac{6r - 4}{4r + 8} $ The denominator of the second term is already $4r + 8$ , so we don't need to change it. To get $4r + 8$ in the denominator of the third term, multiply it by $\frac{2}{2}$ $ \dfrac{10}{2r + 4} \times \dfrac{2}{2} = \dfrac{20}{4r + 8} $ This give us: $ \dfrac{6r - 4}{4r + 8} = \dfrac{1}{4r + 8} + \dfrac{20}{4r + 8} $ If we multiply both sides of the equation by $4r + 8$ , we get: $ 6r - 4 = 1 + 20$ $ 6r - 4 = 21$ $ 6r = 25 $ $ r = \dfrac{25}{6}$